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$26\%$ of adults wear contact lenses and two are selected randomly. Given that at least $1$ of these adults wears lenses, what is the probability that both wear contact lenses?

So I figured $$P(A \text{ and } B) = P(A) P(B|A) = 0.26 \times 0.26 = P(0.26) P(B|A).$$ So I divide $0.0676$ by $0.26$ and get $0.26$. I'm not sure that I've done this right I've been looking for answers online and I've been stuck for an hour. Help is much needed. Thanks.

I also tried adding the sums of the possibilities. Probability of $A$ but not $B$. So $0.26 \times 0.76$ then (Prob of not $A$ but $B$) $+ 0.76 \times 0.26$ which gave me $\longrightarrow 0.1924 + 0.1924 = 0.3848$. Then divide $0.0676$ by $0.3848$? Sorry I don't know how to make proper math symbols.

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  • $\begingroup$ Continue to include your thoughts and efforts in future posts. Formatting tips here. $\endgroup$
    – Em.
    Commented Jan 9, 2016 at 21:40
  • $\begingroup$ It's so tempting to answer $0.26$. $\endgroup$
    – CommonerG
    Commented Jan 9, 2016 at 21:45

1 Answer 1

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26% of adults wear contact lenses and two are selected randomly. Given that at least 1 of these adults wears lenses

Let $A = \{\text{At least 1 wears lenses}\}$, $B = \{\text{2 wear lenses}\}$. Then \begin{align*} P(B|A) &= \frac{P(AB)}{P(A)}\\ &= \frac{P(B)}{P(\text{One of two wears})+P(\text{Two of two wear})} \\ &= \frac{.26(.26)}{\binom{2}{1}(.26)(1-.26)+\binom{2}{2}(.26)^2(1-.26)^0}\\ &= 0.1494253 \end{align*} since $A\cap B =B$.

You can also calculate as follows \begin{align*} P(B|A) &= \frac{P(AB)}{P(A)}\\ &= \frac{P(B)}{1-P(\bar A)} \\ &=\frac{.26(.26)}{1-\binom{2}{0}(.26)^0(1-.26)^2}\\ &=0.1494253. \end{align*}

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  • $\begingroup$ Why do you add the sum of the probability of Two of Two wear? Is it just because you have to take every circumstance then divide? $\endgroup$
    – Taylor Kern
    Commented Jan 9, 2016 at 21:32
  • $\begingroup$ $P(A)$ means the probability of at least one wears lenses. Then either one wears lenses or two wear lenses. So $P(A)$ is equal to the sum of the probability that exactly one wears or exactly two wears, since these events are mutually exclusive. $\endgroup$
    – Em.
    Commented Jan 9, 2016 at 21:39
  • $\begingroup$ Thank you very much for your help! I appreciate it very much. $\endgroup$
    – Taylor Kern
    Commented Jan 9, 2016 at 21:44
  • $\begingroup$ @TaylorKern No problem. Glad to help. $\endgroup$
    – Em.
    Commented Jan 9, 2016 at 21:52

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